Let us prove that (iv) \(\Rightarrow \)  \(d^2=o(n)\) . By Theorem 1 in [1]} (or Theorem 1.1 in [2]}), \({\mathbb {P}}(S_n-n\leqslant 1)\geqslant 1/13\)  for all \(n\) . By standard inequalities for concentration functions (see Theorem 2.22 in [3]}), there exists \(C>0\)  such that \({\mathbb {P}}(1-\lambda \leqslant S_n-n\leqslant 1)\leqslant C(\lambda +1)/\sqrt{n}\)  for all \(\lambda \geqslant 0\)  and \(n\) .
Taking \(\lambda =\gamma \sqrt{n}+1\)  for small enough \(\gamma >0\) , we will guarantee that \({\mathbb {P}}(- \gamma \sqrt{n}\leqslant S_n-n\leqslant 1)\leqslant 1/26\)  for all large enough \(n\) . As a result, we will get that \({\mathbb {P}}( S_n-n\leqslant - \gamma \sqrt{n})\geqslant 1/26+o(1)\) . If (iv) holds, then \( {\mathbb {P}}( S_n-n\leqslant - \varepsilon n/d)\rightarrow 0\)  for any \(\varepsilon >0\) . The latter is possible only if, for any fixed \(\varepsilon >0\) , \(\varepsilon n/d>\gamma \sqrt{n}\)  for all large enough \(n\) , i.e. \(n\geqslant n_0(\varepsilon )\) . The latter means that \(\sqrt{n}=o(n/d)\)  or, equivalently, \(d^2=o(n).\)
